# An inequality of the mean involving truncation

Let ${X_1,X_2,...}$ be i.i.d. r.vs with ${\mathbb{E}[|X_i|] < \infty}$ and ${Y_k = X_k \mathbb{I}_{(|X_k| \leq k)}}$. Then

$\displaystyle \boxed{ \mathbb{E}[X_1] \geq \sum_{k=1}^{\infty} \frac{var(Y_k)}{4 k^2 } }$

Proof:

First we proove the following useful result

If ${X \geq 0}$ and ${ a > 0}$ then

$\displaystyle \boxed{ \mathbb{E}[X^a] = \int_{0}^{\infty} a x^{a-1} P(X >x) dx }$

$\displaystyle \begin{array}{rcl} \int_{0}^{\infty} a x^{a-1} P(X >x) dx &=& \int_{0}^{\infty} \int_{\Omega}a x^{a-1} \mathbb{I}_{(X>x)} dP dx \\ &=& \int_{\Omega} \int_{0}^{\infty} a x^{a-1} \mathbb{I}_{(X>x)} dP dx \\ &=& \int_{\Omega} \int_{0}^{X} a x^{a-1} dP dx = \mathbb{E}[X^a] \end{array}$

Note you can find the same lemma on Feller Vol.2 (p. 150) as

$\displaystyle \mathbb{E}[X^a]= \int_{0}^{\infty} x^a F \{ dx \} = a \int_{0}^{\infty} x^{a-1} [ 1- F(x)] dx$

Using the result above

$\displaystyle \begin{array}{rcl} var(Y_k) \leq \mathbb{E}[Y_k^2] &=& \int_{0}^{\infty}2y P(|Y_k|>y)dy \\ &\leq& \int_{0}^{k}2 y P(|X_1|>y)dy \end{array}$

So

$\displaystyle \begin{array}{rcl} \sum_{k=1}^{\infty} \mathbb{E}[Y_k^2]/k^2 &\leq& \sum_{k=1}^{\infty} k^{-2} \mathbb{I}_{(yy ) dy \\ &=& \int_{0}^{\infty} \left\lbrace \sum_{k=1}^{\infty} k^{-2} \mathbb{I}_{(yy ) dy \end{array}$

Now note that if ${m \geq 2}$

$\displaystyle \sum_{k \geq m} k^{-2} \leq \int_{m-2}^{\infty} x^{-2}dx = \frac{1}{m-1}$

When ${y \geq 1}$, ${k}$ takes the initial value of ${\left\lfloor y \right\rfloor +1 \geq 2}$, so

$\displaystyle 2y \sum_{k > y} k^{-2} \leq 2y / \left\lfloor y \right\rfloor \leq 4$

while when ${ 0 \leq y <1}$,

$\displaystyle 2y \sum_{k > y} k^{-2} \leq 2 \left( 1 + \sum_{k=2}^{\infty} k^{-2} \right) \leq 4$

Hence

$\displaystyle \begin{array}{rcl} \sum_{k=1}^{\infty} \frac{var(Y_k)}{k^2} &\leq& \int_{0}^{\infty} \left\lbrace \sum_{k=1}^{\infty} k^{-2} \mathbb{I}_{(yy ) dy \\ &\leq& \int_{0}^{\infty}4 P(|X_1|>y ) dy \\ &=& 4 \mathbb{E}[|X_1|] \end{array}$